May 2020 - mathfun@mailman.xmission.com

[math-fun] (2,3) and (3,4) Torus Knot Certificates
by Brad Klee 31 May '20

31 May '20

https://0x0.st/ifEP.txt https://0x0.st/ifEX.txt Meta-Data, by (line #): (1) (p,q) torus knot Parametric Eqs, (2) Z-linear surface, degree q, (3) Z-quadratic surface, degree 4, (4) dt/d\phi, (5) Annihilator of dt/du, an ODE in k, (6) Annihilator Certificate, A[dt/du]-d/dt(cert)=0. Imposing Symmetry leaves one degree of freedom. After trying many different constraints, I found that allowing, in fact requiring, an extra point (X,Y,Z)=(0,0,0) leads to smallest known proof data, see linked certificates above. For the first two cases (p,q) the Z-linear surface had degree q, and the ODE has q terms. This suggests apparent genus 1 and 2 for the first two torus knots, (2,3) and (3,4). When measuring genus by period ODE, it often happens that apparent genus is less than real genus due to hidden symmetry. I don't know about slice genus, but according to the proved Milnor conjecture, the numbers are 1 and 3, see corollary 1 of: https://arxiv.org/abs/math/0402131 I doubt that any representation of the stevedore knot would have a rational period function, but it is Lissajous, so it might possible to check. Unfortunately, I don't have anymore time this week. --Brad

1 0

Re: [math-fun] Knot Problem
by Steve Witham 31 May '20

31 May '20

This has precious* little to do with Dan's method, but he spurred me locate the Grapher app on my computer (Utilities folder??), take a demo that comes with it (Torus-knot), turn it into a trefoil knot, and then fix the graphics that got broke. This page shows the settings as well as the picture: http://www.mac-guyver.com/switham/2020/05/Torus_knot <http://www.mac-guyver.com/switham/2020/05/Torus_knot/index.html> (As far as I know, the only way to get equations into Grapher is with its own editor. Their file format looks opaque. It can apparently import point sets, though...) --Steve *my precious. Fit for an elven king. Btw, summary of _The Hobbit_: Marty Feldman gets hold of an all-access key card to Cheyenne Mountain. > Dan Asimov <dasimov(a)earthlink.net> > Date: 5/25/20, 5:45 PM > > I tried to find equations in R^3 for the trefoil defined initially in > R^4 = {(x,y,z,w)}: > > (x+iy)^2 + (z+iw)^3 = 0 > > and > > x^2 + y^2 + z^2 + w^2 = 1 > > but which is then stereographically projected into > R^3 = {(X,Y,Z)} via > > X = x/(1-w), Y = y/(1-w), Z = z/(1-w). > > If my algebra is right, the final (two) equations are > > 4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 > & > 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0 > > where W is short for (X^2 + Y^2 + Z^2 - 1). > > I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher". > > Â—Dan > >

1 0

Re: [math-fun] "Wrong Times Table"
by Dan Asimov 30 May '20

30 May '20

Who dat? —Dan ----- Joe Soap ... -----

2 1

[math-fun] Different graph question
by Keith F. Lynch 30 May '20

30 May '20

Allan Wechsler <acwacw(a)gmail.com> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> Find a connected graph of girth 5 (i.e. no loops of size 3 or 4) in >> which every integer is a vertex, every vertex has the same valence, >> and every vertex is part of a loop of size 5. >> Yes, I have a solution, unless I'm confused. > All even numbers 2n are connected to 2n+1, 2n+2, and 2n-2. All odd > numbers 2n+1 are connected to 2n, 2n+5, and 2n-4. I am pretty sure > this graph is cubic of girth 5. 3 is connected to 7, but 7 is not connected to 3.

2 1

Re: [math-fun] Different graph question
by Dan Asimov 30 May '20

30 May '20

As long as (p-2)(q-2) > 4 there exists a regular tiling of the hyperbolic plane H by regular hyperbolic p-gons that meet q per vertex. Suppose we have such a tiling with p = 5 and q >= 3. Then the union of all the edges of this tiling form a graph with countably many vertices and constant valence = q; label them using any bijection Z <—> {vertices}. —Dan _____ * Proof sketch: Think of the Poincaré disk model of H. Start with an arbitrarily small regular p-gon, whose interior angles will be arbitrarily close to Euclidean, i.e. (p-2)π/p, and slowly enlarge it. The angles will decrease toward 0 since geodesics meet the boundary of the unit disk at right angles. Stop when they are 2π/q. Since H is the same everywhere, the q tiles per vertex tiling can be continued indefinitely. ----- Allan Wechsler <acwacw(a)gmail.com> wrote: Find a connected graph of girth 5 (i.e. no loops of size 3 or 4) in which every integer is a vertex, every vertex has the same valence, and every vertex is part of a loop of size 5. -----

1 0

[math-fun] "Wrong Times Table"
by Keith F. Lynch 30 May '20

30 May '20

Is there anything mathematically interesting about today's XKCD, https://xkcd.com/2313, "Wrong Times Table"? It's commutative, or at least no exceptions can be found. But it is *not* associative. There are 48 ways to test associativity, and it passes exactly 24 and flunks exactly 24, which is slightly interesting. None of the rows or columns is monotonic. Many rows and columns have duplicate numbers. Not all the entries are integers. There's a zero divisor.

4 3

[math-fun] Fun with "digit-average" (long post)
by Éric Angelini 30 May '20

30 May '20

Hello Math-Fun, The idea, here, is to play with the notion of "digit-average". If we take the set {0,1,2,3,4,5,6,7,8,9}, make the sum and divide by the number of elements, we get 45/10 which gives 4.5 as the average value of a digit belonging to the set. What about re-inserting this value back in the set? We would then have the new set {0,1,2,3,4,4,5,5,6,7,8,9} with sum 54 for 12 elements and "digit-average" 54/12 = 4.5 (the same as before). Re-inserting this value would produce the 3rd set {0,1,2,3,4,4,4,5,5,5,6,7,8,9} with sum 63 for 14 elements and "digit-average" of 63/14 = 4.5 again. The successive sets show an obvious pattern. Instead of {0,1,2,3,4,5,6,7,8,9}, let's start with {1,3}. Sum is 4 with 2 digits; average = 2; new set is {1,2,3}; sum is 6 with 3 digits; average = 2; new set is {1,2,2,3}, etc. Similar pattern. But this gets weird sometimes. Start with {1,10}. Sum is 2 for 3 digits; average = 0.66666666... How do we plug this value back in the set? Well, let's decide that we dont plug any repeated block of digits, only the first one (this is the "truncation rule"). We would then have here as 2nd set {0,1,6,10}; sum 8 with 5 digits; average = 1.6; the 3rd set would now be {0,1,1,6,6,10}; sum 15 with 7 digits; average = 2.142857142857142857... According to the truncation rule, we'll plug back in the set the value 2.142857. The 4th set will then look like {0,1,1,1,2,2,4,5,6,6,7,8,10}. This notation is confusing because we don't see easily what has been added exactly to the former set. Let's decide to not reorder the elements and to see the new set as an extended sequence. We would then have S: S = 1, 10, 0, 6, 1, 6, 2, 142857,... This is clearer and shows us three things: a) the start of S is given by a(1) and a(2); b) the sequence develops all by itself, with no exterior intervention (the truncation rule is enough); c) the successive pairs of terms after a(1) and a(2) form the successive "digit-averages" that were plugged back into the sequence (truncated or not) – and this is nice to recognize: S = 1, 10, 0, 6, 1, 6, 2, 142857,... The question I was asking myself yesterday was: which S's will enter into a loop at some point? (...) The end of this long post is here, on my personal blog: http://cinquantesignes.blogspot.com/2020/05/digits-average.html Best, É.

1 0

[math-fun] Some math puzzles
by Dan Asimov 29 May '20

29 May '20

Here are four problems from the puzzle column in the latest newsletter from MSRI, faintly edited. —Dan ----- 1. Find all possible positive integers a, b, c, and d such that ab = c+d and cd = a+b. 2. Let C be the surface of a cube in 3-space. What is the largest n such that there is a planar regular n-gon P, not lying entirely on one face of C, such that all P’s vertices are on C? 3. Suppose there are n hats, each with a different color. These hats are placed on the heads of n sages. All of the sages know all the colors: their own hat and everyone else’s hats. A referee then announces the "correct" hat color that should be on the head of each sage. The sages are then allowed "swap" sessions: in one session disjoint pairs of sages are allowed to interchange their hats. Can the sages fully correct their hat colors in two swap sessions? 6. N people are randomly placed on a circle. Each person turns to face their closest neighbor (with probability 1, all the distances are distinct, so this is sufficiently well-defined). What is the expected number of "lonely people" who are looking at a neighbor who is not looking at them? -----

1 0

[math-fun] Trigonopeltastes delta
by Brad Klee 29 May '20

29 May '20

https://www.inaturalist.org/observations/47750517 (pupated and hatched in captivity, surprised me!) The common name is "delta flower scarab". But the yellow triangle on the pronotum is upside down relative to Greek delta, so I call them "nabla flower scarabs" instead. That is okay because common names are not standardized, nor unique. The nabla flower scarab is found May through August in much of the S.E. United states. Members of this species like to spend their time calculating the normal vectors of flower petals. --Brad

1 0

Re: [math-fun] prob gen fcn for *product* of Bernoulli random vars?
by Dan Asimov 29 May '20

29 May '20

The software called "pari" will factor polyomials with rational coefficients into irreducible factors with rational coefficients. —Dan ----- If I have a Bernoulli random variable with a probability p of value 2 and a probability q=(1-p) of value 98, I can represent its PGF as G(z,p,q,1), see below. Is G(z,p,q,n) below correct for the PGF of the *product* of n independent such random samples? (%i1) G(z,p,q,1); 98 2 (%o1) q z + p z (%i2) G(z,p,q,2); 2 9604 196 2 4 (%o2) q z + 2 p q z + p z (%i3) G(z,p,q,n); n ==== i n - i \ i n - i 2 98 (%o3) > binomial(n, i) p q z / ==== i = 0 --- An aside: trying to *factor* such a polynomial as in %o2 above causes mucho pain! Are there any poly factoring algs that can factor polys like these (or prove they aren't factorable) ? -----

1 0